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NCERT Class 10 Mathematics Exercise 8.3 Solutions: Introduction to Trigonometry
AI-written, checked on 23 Oct 2026
Solutions written by Shishya, checked — not the board's official answers
Step-by-step working for every question of Exercise 8.3 in NCERT's Class 10 Mathematics book, chapter “Introduction to Trigonometry”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.
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Ex 8.3 Q1
Write sin A, sec A and tan A in terms of cot A.
- We know that sin A and cot A are related by the identity sin A = 1 / cosec A, and cosec A = √(1 + cot² A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Therefore, sin A = 1 / √(1 + cot² A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- We know that sec A and cot A are related by the identity sec A = √(1 + tan² A), and tan A = 1 / cot A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Substituting tan A, we get sec A = √(1 + (1 / cot A)²).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Simplifying the expression inside the square root: sec A = √((cot² A + 1) / cot² A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Taking the square root of the numerator and denominator separately: sec A = √(1 + cot² A) / cot A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- We know that tan A is the reciprocal of cot A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Therefore, tan A = 1 / cot A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
Answer: sin A = 1 / √(1 + cot² A); sec A = √(1 + cot² A) / cot A; tan A = 1 / cot A
- We know that sin A and cot A are related by the identity sin A = 1 / cosec A, and cosec A = √(1 + cot² A).
Ex 8.3 Q2
Write all the other trigonometric ratios of A in terms of sec A.
- We know that cos A = 1 / sec A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Using the identity sin² A + cos² A = 1, we get sin² A = 1 − cos² A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Substitute cos A: sin² A = 1 − (1 / sec A)² = 1 − 1 / sec² A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Simplify: sin² A = (sec² A − 1) / sec² A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Take square root: sin A = √(sec² A − 1) / sec A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Now tan A = sin A / cos A = [√(sec² A − 1) / sec A] / [1 / sec A].
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Simplify tan A: tan A = √(sec² A − 1).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- cosec A = 1 / sin A = sec A / √(sec² A − 1).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- cot A = 1 / tan A = 1 / √(sec² A − 1).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
Answer: sin A = √(sec² A − 1) / sec A; tan A = √(sec² A − 1); cosec A = sec A / √(sec² A − 1); cot A = 1 / √(sec² A − 1)
- We know that cos A = 1 / sec A.
Ex 8.3 Q3
Choose and justify: (i) 9 sec²A − 9 tan²A = 1, 9, 8 or 0? (ii) (1 + tan θ + sec θ)(1 + cot θ − cosec θ) = 0, 1, 2 or −1? (iii) (sec A + tan A)(1 − sin A) = sec A, sin A, cosec A or cos A? (iv) (1 + tan²A)/(1 + cot²A) = sec²A, −1, cot²A or tan²A?
- (i) Use the identity 1 + tan²A = sec²A, which implies sec²A − tan²A = 1.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (i) Factor out 9 from the expression: 9(sec²A − tan²A) = 9(1) = 9.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (ii) Let the expression be E = (1 + tan θ + sec θ)(1 + cot θ − cosec θ).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (ii) Convert to sine and cosine: E = (1 + sin θ/cos θ + 1/cos θ)(1 + cos θ/sin θ − 1/sin θ).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (ii) Simplify the first bracket: (cos θ + sin θ + 1)/cos θ.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (ii) Simplify the second bracket: (sin θ + cos θ − 1)/sin θ.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (ii) Multiply the fractions: E = [(cos θ + sin θ + 1)(cos θ + sin θ − 1)] / (sin θ cos θ).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (ii) Apply difference of squares in the numerator: [(cos θ + sin θ)² − 1²] / (sin θ cos θ).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (ii) Expand the square: [cos²θ + sin²θ + 2 sin θ cos θ − 1] / (sin θ cos θ).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (ii) Use cos²θ + sin²θ = 1 to cancel terms: [1 + 2 sin θ cos θ − 1] / (sin θ cos θ) = 2 sin θ cos θ / (sin θ cos θ) = 2.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (iii) Let the expression be F = (sec A + tan A)(1 − sin A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (iii) Convert to sine and cosine: F = (1/cos A + sin A/cos A)(1 − sin A) = [(1 + sin A)/cos A](1 − sin A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- (iii) Multiply the terms: F = (1 + sin A)(1 − sin A) / cos A = (1 − sin²A) / cos A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (iii) Use 1 − sin²A = cos²A: F = cos²A / cos A = cos A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (iv) Let the expression be G = (1 + tan²A) / (1 + cot²A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (iv) Use identities 1 + tan²A = sec²A and 1 + cot²A = cosec²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (iv) Substitute: G = sec²A / cosec²A = (1/cos²A) / (1/sin²A) = sin²A / cos²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- (iv) Simplify to tan²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
Answer: (i) (B); (ii) (C); (iii) (D); (iv) (D)
- (i) Use the identity 1 + tan²A = sec²A, which implies sec²A − tan²A = 1.
Ex 8.3 Q4(i)
Prove (acute angle, expressions defined): (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ).
- Start with the left-hand side: (cosec θ − cot θ)².
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Express cosec θ and cot θ in terms of sin θ and cos θ: (1/sin θ − cos θ/sin θ)².
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Combine the fractions inside the square: ((1 − cos θ)/sin θ)².
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Square the numerator and denominator: (1 − cos θ)² / sin² θ.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity sin² θ = 1 − cos² θ to replace the denominator: (1 − cos θ)² / (1 − cos² θ).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Factor the denominator as a difference of squares: (1 − cos θ)² / [(1 − cos θ)(1 + cos θ)].
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Cancel the common factor (1 − cos θ) from numerator and denominator: (1 − cos θ) / (1 + cos θ).
- This matches the right-hand side, so the identity is proved.
Answer: (i) Proved: (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ)
- Start with the left-hand side: (cosec θ − cot θ)².
Ex 8.3 Q4(ii)
Prove (acute angle, expressions defined): cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.
- Take the least common denominator for the left-hand side, which is cos A (1 + sin A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Combine the fractions to get [cos² A + (1 + sin A)²] / [cos A (1 + sin A)].
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Expand the numerator to cos² A + 1 + 2 sin A + sin² A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity sin² A + cos² A = 1 to simplify the numerator to 2 + 2 sin A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Factor out 2 from the numerator to get 2(1 + sin A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Cancel the common factor (1 + sin A) from the numerator and denominator.
- The expression simplifies to 2 / cos A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Since 1 / cos A is defined as sec A, the result is 2 sec A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Thus, the left-hand side equals the right-hand side, proving the identity.
Answer: Proved: cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
- Take the least common denominator for the left-hand side, which is cos A (1 + sin A).
Ex 8.3 Q4(iii)
Prove (acute angle, expressions defined): tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ.
- Let LHS = tan θ/(1 − cot θ) + cot θ/(1 − tan θ).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute tan θ = sin θ/cos θ and cot θ = cos θ/sin θ into the expression.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Simplify the first term: (sin θ/cos θ) / (1 − cos θ/sin θ) = (sin θ/cos θ) / ((sin θ − cos θ)/sin θ) = sin²θ / (cos θ(sin θ − cos θ)).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Simplify the second term: (cos θ/sin θ) / (1 − sin θ/cos θ) = (cos θ/sin θ) / ((cos θ − sin θ)/cos θ) = cos²θ / (sin θ(cos θ − sin θ)).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Rewrite the second term with a common denominator by noting (cos θ − sin θ) = −(sin θ − cos θ), so the term becomes −cos²θ / (sin θ(sin θ − cos θ)).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Combine the terms over the common denominator sin θ cos θ (sin θ − cos θ): [sin³θ − cos³θ] / [sin θ cos θ (sin θ − cos θ)].
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Factor the numerator using the identity a³ − b³ = (a − b)(a² + ab + b²) to get (sin θ − cos θ)(sin²θ + sin θ cos θ + cos²θ) / [sin θ cos θ (sin θ − cos θ)].
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Cancel the common factor (sin θ − cos θ) to obtain (sin²θ + sin θ cos θ + cos²θ) / (sin θ cos θ).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity sin²θ + cos²θ = 1 to simplify the numerator to 1 + sin θ cos θ.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Split the fraction: 1/(sin θ cos θ) + (sin θ cos θ)/(sin θ cos θ) = 1/(sin θ cos θ) + 1.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Recognize that 1/(sin θ cos θ) = (1/sin θ)(1/cos θ) = cosec θ sec θ.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Thus, LHS = 1 + sec θ cosec θ, which equals the RHS.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Hence, the identity is proved.
Answer: Proved: tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ
- Let LHS = tan θ/(1 − cot θ) + cot θ/(1 − tan θ).
Ex 8.3 Q4(iv)
Prove (acute angle, expressions defined): (1 + sec A)/sec A = sin²A/(1 − cos A).
- Start with the left-hand side: (1 + sec A) / sec A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Divide both terms in the numerator by sec A to get 1 + 1/sec A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute 1/sec A with cos A, so the LHS becomes 1 + cos A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Now consider the right-hand side: sin²A / (1 − cos A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity sin²A = 1 − cos²A to replace the numerator.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Factor the numerator as (1 − cos A)(1 + cos A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- The expression becomes (1 − cos A)(1 + cos A) / (1 − cos A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Cancel the common factor (1 − cos A) from the numerator and denominator.
- The RHS simplifies to 1 + cos A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Since LHS = RHS, the identity is proved.
Answer: Proved: (1 + sec A)/sec A = sin²A/(1 − cos A)
- Start with the left-hand side: (1 + sec A) / sec A.
Ex 8.3 Q4(v)
Prove (acute angle, expressions defined): (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A, using cosec²A = 1 + cot²A.
- Divide the numerator and the denominator of the left-hand side by sin A: LHS = (cot A − 1 + cosec A)/(cot A + 1 − cosec A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Replace the 1 in the numerator by cosec²A − cot²A (since cosec²A = 1 + cot²A): numerator = (cot A + cosec A) − (cosec²A − cot²A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Factorise cosec²A − cot²A = (cosec A + cot A)(cosec A − cot A), so numerator = (cosec A + cot A)[1 − (cosec A − cot A)] = (cosec A + cot A)(1 − cosec A + cot A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- The denominator is cot A + 1 − cosec A = 1 − cosec A + cot A, the same bracket.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Cancel it: LHS = cosec A + cot A = RHS.
Answer: Proved: (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A
- Divide the numerator and the denominator of the left-hand side by sin A: LHS = (cot A − 1 + cosec A)/(cot A + 1 − cosec A).
Ex 8.3 Q4(vi)
Prove (acute angle, expressions defined): √[(1 + sin A)/(1 − sin A)] = sec A + tan A.
- Start with the Left Hand Side: √[(1 + sin A)/(1 − sin A)].
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Multiply the numerator and denominator inside the square root by (1 + sin A) to rationalize the denominator.
why?
Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.
- The expression becomes √[(1 + sin A)² / (1 − sin A)(1 + sin A)].
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Use the identity 1 − sin²A = cos²A in the denominator to get √[(1 + sin A)² / cos²A].
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Take the square root of the numerator and denominator separately: (1 + sin A) / cos A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Split the fraction into two terms: 1/cos A + sin A/cos A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Use the definitions sec A = 1/cos A and tan A = sin A/cos A to rewrite the expression as sec A + tan A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- This matches the Right Hand Side, hence the identity is proved.
Answer: Proved: √[(1 + sin A)/(1 − sin A)] = sec A + tan A
- Start with the Left Hand Side: √[(1 + sin A)/(1 − sin A)].
Ex 8.3 Q4(vii)
Prove (acute angle, expressions defined): (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ.
- Take the left-hand side: (sin θ − 2 sin³θ) / (2 cos³θ − cos θ)
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Factor sin θ from the numerator: sin θ (1 − 2 sin²θ)
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Factor cos θ from the denominator: cos θ (2 cos²θ − 1)
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity sin²θ + cos²θ = 1 to write 1 − 2 sin²θ as 1 − 2(1 − cos²θ) = 2 cos²θ − 1
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Substitute this into the numerator: sin θ (2 cos²θ − 1)
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- The expression becomes [sin θ (2 cos²θ − 1)] / [cos θ (2 cos²θ − 1)]
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Cancel the common factor (2 cos²θ − 1) from numerator and denominator
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- The remaining expression is sin θ / cos θ
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- By definition, sin θ / cos θ = tan θ
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Hence, the left-hand side equals the right-hand side, proving the identity
Answer: Proved: (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ
- Take the left-hand side: (sin θ − 2 sin³θ) / (2 cos³θ − cos θ)
Ex 8.3 Q4(viii)
Prove (acute angle, expressions defined): (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A.
- Expand the left-hand side: (sin A + cosec A)² + (cos A + sec A)² = sin²A + 2 sin A cosec A + cosec²A + cos²A + 2 cos A sec A + sec²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Simplify the cross terms using sin A cosec A = 1 and cos A sec A = 1, giving sin²A + 2 + cosec²A + cos²A + 2 + sec²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Group the squared terms: (sin²A + cos²A) + (cosec²A + sec²A) + 4.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Apply the identity sin²A + cos²A = 1, resulting in 1 + cosec²A + sec²A + 4 = 5 + cosec²A + sec²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identities cosec²A = 1 + cot²A and sec²A = 1 + tan²A to substitute into the expression.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Substitute to get 5 + (1 + cot²A) + (1 + tan²A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Combine the constants: 5 + 1 + 1 = 7, leaving 7 + tan²A + cot²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- This matches the right-hand side, proving the identity.
Answer: Proved: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
- Expand the left-hand side: (sin A + cosec A)² + (cos A + sec A)² = sin²A + 2 sin A cosec A + cosec²A + cos²A + 2 cos A sec A + sec²A.
Ex 8.3 Q4(ix)
Prove (acute angle, expressions defined): (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A).
- Take the left-hand side: (cosec A − sin A)(sec A − cos A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Replace cosec A with 1/sin A and sec A with 1/cos A to get (1/sin A − sin A)(1/cos A − cos A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Simplify the first bracket: (1 − sin² A)/sin A = cos² A/sin A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Simplify the second bracket: (1 − cos² A)/cos A = sin² A/cos A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Multiply the simplified brackets: (cos² A/sin A) × (sin² A/cos A) = cos A sin A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Take the right-hand side: 1/(tan A + cot A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Replace tan A with sin A/cos A and cot A with cos A/sin A to get 1/(sin A/cos A + cos A/sin A).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Combine the terms in the denominator: (sin² A + cos² A)/(sin A cos A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity sin² A + cos² A = 1 to simplify the denominator to 1/(sin A cos A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- The right-hand side becomes 1 / (1/(sin A cos A)) = sin A cos A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Since the left-hand side equals sin A cos A and the right-hand side equals sin A cos A, LHS = RHS.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
Answer: Proved
- Take the left-hand side: (cosec A − sin A)(sec A − cos A).
Ex 8.3 Q4(x)
Prove (acute angle, expressions defined): (1 + tan²A)/(1 + cot²A) = [(1 − tan A)/(1 − cot A)]² = tan²A.
- Consider the left-hand side: (1 + tan²A)/(1 + cot²A).
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity 1 + tan²A = sec²A to replace the numerator.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Use the identity 1 + cot²A = cosec²A to replace the denominator.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- The expression becomes sec²A / cosec²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Write sec²A as 1/cos²A and cosec²A as 1/sin²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Simplify the fraction to (1/cos²A) × (sin²A/1) = tan²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Now consider the middle expression: [(1 − tan A)/(1 − cot A)]².
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Write tan A as sin A/cos A and cot A as cos A/sin A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute these into the numerator: 1 − sin A/cos A = (cos A − sin A)/cos A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute these into the denominator: 1 − cos A/sin A = (sin A − cos A)/sin A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Divide the numerator by the denominator: [(cos A − sin A)/cos A] / [(sin A − cos A)/sin A].
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Note that cos A − sin A = −(sin A − cos A), so the ratio becomes −sin A/cos A = −tan A.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Square the result: (−tan A)² = tan²A.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
- Since both the left-hand side and the middle expression equal tan²A, the given equation is proved.
why?
sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.
Answer: (1 + tan²A)/(1 + cot²A) = [(1 − tan A)/(1 − cot A)]² = tan²A
- Consider the left-hand side: (1 + tan²A)/(1 + cot²A).
Other exercises in this chapter
- Exercise 8.1 — 11 questions
- Exercise 8.2 — 4 questions
← Introduction to Trigonometry: notes, practice and all exercises
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